3Cl2 +6KOH->KClO3 +5KCl =3H2O
write a conversion factor that could be used to convert between moles od potassium hydroxide and moles of potassium chlorate
You have a typo.
3Cl2 +6KOH->KClO3 +5KCl + 3H2O
z mols KOH x (1 mol KClO3/5 mols KOH) = ? mols KClO3
To convert between moles of potassium hydroxide (KOH) and moles of potassium chlorate (KClO3), we can use the balanced equation:
3Cl2 + 6KOH → KClO3 + 5KCl + 3H2O
From this balanced equation, we can see that the stoichiometric ratio between KOH and KClO3 is 6:1. That means for every 6 moles of KOH consumed, 1 mole of KClO3 is produced.
Therefore, the conversion factor is:
1 mole KClO3 / 6 moles KOH
To convert between moles of potassium hydroxide (KOH) and moles of potassium chlorate (KClO3), we can use the stoichiometric coefficients from the balanced chemical equation.
From the equation 3Cl2 + 6KOH -> KClO3 + 5KCl + 3H2O, we can see that the molar ratio between KOH and KClO3 is 6:1.
Therefore, the conversion factor for converting from moles of KOH to moles of KClO3 is:
1 mol KClO3 / 6 mol KOH
This conversion factor tells us that for every 6 moles of KOH, we will have 1 mole of KClO3.